Quadratic Equations
Condition for quadratic to be non-negative
Grade 11
Question:
<p>If <span>\( f(x) \geq -3 \)</span> for all <span>\( x \in \mathbb{R} \)</span>, and the inequality <span>\( x^2 + 2px + 4p + f(x) \geq 0 \)</span> holds for all <span>\( x \in \mathbb{R} \)</span>, find the range of <span>\( p \)</span> such that <span>\( p^2 - 4p + 3 < 0 \)</span>, i.e., <span>\( 1 \leq p \leq 3 \)</span>. How many integer values does <span>\( p \)</span> take?</p>
Step-by-Step Solution
Key Concept: Since f(x) ≥ -3 for all x ∈ ℝ, the constraint x² + 2px + 4p + f(x) ≥ 0 for all x is most restrictive when f(x) achieves its minimum value of -3. This forces x² + 2px + 4p - 3 ≥ 0 for all x, requiring the discriminant condition 4p² - 4(4p - 3) ≤ 0, which simplifies to p² - 4p + 3 ≤ 0.
<p><strong>Step 1:</strong> Identify the constraint on f(x): Given f(x) ≥ -3 for all x ∈ ℝ, the minimum value of f(x) is -3.</p><p><strong>Step 2:</strong> For x² + 2px + 4p + f(x) ≥ 0 to hold for all x ∈ ℝ, it must hold even when f(x) is at its minimum. Thus: x² + 2px + 4p - 3 ≥ 0 for all x ∈ ℝ.</p><p><strong>Step 3:</strong> A quadratic ax² + bx + c ≥ 0 for all x requires discriminant Δ ≤ 0. Here: Δ = (2p)² - 4(1)(4p - 3) ≤ 0</p><p><strong>Step 4:</strong> Simplify: 4p² - 16p + 12 ≤ 0 → p² - 4p + 3 ≤ 0 → (p - 1)(p - 3) ≤ 0</p><p><strong>Step 5:</strong> Solving the inequality: 1 ≤ p ≤ 3</p><p><strong>Step 6:</strong> Count integers in [1, 3]: p ∈ {1, 2, 3}</p><p>∴ Answer: <strong>3 integer values</strong></p>
Correct Answer: 6